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6 February, 05:30

Two wires with equal lengths are made of pure copper. The diameter of wire A is three times the diameter of wire B. When 8 kg masses are hung on the wires, wire B stretches more than wire A. You make careful measurements and compute Young's modulus for both wires. What do you find? a. YA > YBb. YA = YBc. YA < YB

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  1. 6 February, 06:01
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    Answer: c. YA < YB

    Explanation:

    The formula for Young's modulus is = Tensile stress / Tensile strain

    Tensile stress = Force x Length

    Force = mass x acceleration due to gravity

    = 8kg x 10m/s

    = 80kgm/s

    Tensile stress = 80kgm/s x 2m = 160kgm2/s

    Tensile strain = Area x change in length

    Area = pi x D2 / 4; Pi = 3.14

    Change in length = L2 - L1 (New length - Initial length)

    Given parameters:

    Length of wire A = Length of wire B, (let's use 2meters for the calculation)

    For wire A, Diameter = 3 x Wire B diameter

    Assuming Diameter of wire B = 1meter

    Therefore, diameter of wire A = 1 x 3 = 3meters

    It is said that wire B stretches more than wire A when the man of 8kg is placed on both

    For wire B, let's assume new length is = 4m

    For wire A let's assume new length is = 3m.

    (i) Tensile strain of wire A =

    Area of wire A = 3.14 x (32) / 4 = 7.065m2

    Change in length = 3m - 2m = 1m.

    Therefore, tensile strain = 7.065m2 x 1m = 7.065m3

    Young's modulus for wire A (YA) = 160kgm2/s divided by 7.065m3

    = 22.64Pa.

    (ii) Tensile strain of wire B =

    Area of wire B = 3.14 x (12) / 4 = 0.785m2

    Change in length = 4m - 2m = 2m

    Therefore, tensile strain = 0.785m2 x 2m = 1.57m3

    Young's modulus for wire B (YB) = 160kgm2/s divided by 1.57m3

    = 101.91Pa.

    From the calculations above, we see that YA is less than YB (YA < YB). This is true given that wire A has a greater diameter than wire B which in turn impacts the Area of the wire since the diameter is directly proportional to area and the area is inversely proportional to the young's modulus.
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